Question

If $PQ$ is a double ordinate of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $OPQ$ is an equilateral triangle, $O$ being the centre of the hyperbola. Then, the eccentricity $e$ of the hyperbola satisfies

• A. $1<e<2/\sqrt{3}$
• B. $e=2/\sqrt{3}$
• C. $e=\sqrt{3}/2$
• D. $e>2/\sqrt{3}$

Answer 1

Answer: (D)

Let $P(a\sec \theta ,b\tan \theta );Q(a\sec \theta ,-b\tan \theta )$ be end points of double ordinate and $C(0,0)$, is the centre of the hyperbola. Now,
$PQ=2b\tan \theta$
$CQ=CP=\sqrt{a^2{\sec }^2\theta +b^2{\tan }^2\theta }$
since $CQ=CP=PQ$
$\therefore 4b^2{\tan }^2\theta =a^2{\sec }^2\theta +b^2{\tan }^2\theta$
$\Rightarrow 3b^2{\tan }^2\theta =a^2{\sec }^2\theta$

$\Rightarrow 3b^2{\sin }^2\theta =a^2$
$\Rightarrow 3a^2\left(e^2-1\right){\sin }^2\theta =a^2$
$\Rightarrow 3\left(e^2-1\right){\sin }^2\theta =1$
$\Rightarrow \frac{1}{3\left(e^2-1\right)}={\sin }^2\theta <1$
$(\because {\sin }^2\theta <1)$
$\Rightarrow \frac{1}{e^2-1}<3\Rightarrow e^2-1>\frac{1}{3}$